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Llana [10]
2 years ago
7

Trials in an experiment with a polygraph include 98 results that include 24 cases of wrong results and 74 cases of correct resul

ts. Use a 0.05 significance level to test the claim that such polygraph results are correct less than 80​% of the time. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.
Mathematics
1 answer:
madreJ [45]2 years ago
5 0

Answer:

z=\frac{0.755 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-1.114  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of successes is not significantly less than 0.8 or 80%

Step-by-step explanation:

Data given and notation

n=98 represent the random sample taken

X=74 represent the number of cases correct

\hat p=\frac{74}{98}=0.755 estimated proportion of successes

p_o=0.8 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of successes is less than 0.8.:  

Null hypothesis:p\geq 0.8  

Alternative hypothesis:p < 0.8  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.755 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-1.114  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of successes is not significantly less than 0.8 or 80%

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